Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $y = \dfrac{3}{7(2k - 3)} \times \dfrac{10(2k - 3)}{5} $
When multiplying fractions, we multiply the numerators and the denominators. $y = \dfrac{ 3 \times 10(2k - 3) } { 7(2k - 3) \times 5 } $ $ y = \dfrac{30(2k - 3)}{35(2k - 3)} $ We can cancel the $2k - 3$ so long as $2k - 3 \neq 0$ Therefore $k \neq \dfrac{3}{2}$ $y = \dfrac{30 \cancel{(2k - 3})}{35 \cancel{(2k - 3)}} = \dfrac{30}{35} = \dfrac{6}{7} $